How do you differentiate #y= (5x)/((tanx)(cotx))#?

3 Answers
Apr 13, 2015

5

tan x * cot x is simply 1, because cot x equals #1/tan x#. Thus it is y= 5x. Hence #dy/dx# =5

Apr 13, 2015

If you don't think first (before you start) you'll use the quotient rule (with the product rule to differentiate the denominator).

Take a few seconds to think, and to ask:
Can I rewrite this before I differentiate to make my life easier?

#tanx = 1/cotx# so #tanx cotx =1#

That means that #y = 5x/1=5x#

So, #y'=5#

Apr 13, 2015

#y'=5#

#y=(5x)/((tanx)(cotx)#

As

#tanx=sinx/cosx#

And

#cotx=cosx/sinx#

So,

#y=(5x)/((sinx/cosx)(cosx/sinx))#

#y=(5x)/((cancelsinx/cancelcosx)(cancelcosx/cancelsinx))#

#y=(5x)/1#

#y=(5x)#

Differentiating both sides with respect to 'x'

#y'=5#